Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}6x-y &= -3 \\ -6x-y &= -9\end{align*}$
We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $1$ $\begin{align*}-6x+y &= 3\\ -6x-y &= -9\end{align*}$ Add the top and bottom equations. $-12x = -6$ Divide both sides by $-12$ and reduce as necessary. $x = \dfrac{1}{2}$ Substitute $\dfrac{1}{2}$ for $x$ in the top equation. $6( \dfrac{1}{2})-y = -3$ $3-y = -3$ $-y = -6$ $y = 6$ The solution is $\enspace x = \dfrac{1}{2}, \enspace y = 6$.